$\dfrac{\tan^2\theta + 1}{\tan^2\theta} = \; ?$
Answer: We can derive a useful identity from ${\sin^2 \theta} + {\cos^2 \theta} = 1$ to simplify this expression. $1$ ${\sin\theta}$ ${\cos\theta}$ $\theta$ We can see why this identity is true by using the Pythagorean Theorem. Dividing both sides by $\cos^2\theta$ , we get $ \dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}$ $ \tan^2\theta + 1 = \sec^2\theta$ Plugging into our expression, we get $ \dfrac{\tan^2\theta + 1}{\tan^2\theta} = \dfrac{\sec^2\theta}{\tan^2\theta} $ To make simplifying easier, let's put everything in terms of $\sin$ and $\cos$ . We know $\sec^2\theta = \frac{1}{\cos^2\theta}$ and $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ , so we can substitute to get $ \dfrac{\sec^2\theta}{\tan^2\theta} = \dfrac{\frac{1}{\cos^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta}} $ This is $\dfrac{1}{\sin^2\theta} = \csc^2\theta$.